3.9 \(\int x^6 \tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}}) \, dx\)

Optimal. Leaf size=114 \[ -\frac{d^2 \left (d+e x^2\right )^{3/2}}{7 e^{7/2}}+\frac{d^3 \sqrt{d+e x^2}}{7 e^{7/2}}-\frac{\left (d+e x^2\right )^{7/2}}{49 e^{7/2}}+\frac{3 d \left (d+e x^2\right )^{5/2}}{35 e^{7/2}}+\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

[Out]

(d^3*Sqrt[d + e*x^2])/(7*e^(7/2)) - (d^2*(d + e*x^2)^(3/2))/(7*e^(7/2)) + (3*d*(d + e*x^2)^(5/2))/(35*e^(7/2))
- (d + e*x^2)^(7/2)/(49*e^(7/2)) + (x^7*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/7

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Rubi [A]  time = 0.0639015, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6221, 266, 43} \[ -\frac{d^2 \left (d+e x^2\right )^{3/2}}{7 e^{7/2}}+\frac{d^3 \sqrt{d+e x^2}}{7 e^{7/2}}-\frac{\left (d+e x^2\right )^{7/2}}{49 e^{7/2}}+\frac{3 d \left (d+e x^2\right )^{5/2}}{35 e^{7/2}}+\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^6*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(d^3*Sqrt[d + e*x^2])/(7*e^(7/2)) - (d^2*(d + e*x^2)^(3/2))/(7*e^(7/2)) + (3*d*(d + e*x^2)^(5/2))/(35*e^(7/2))
- (d + e*x^2)^(7/2)/(49*e^(7/2)) + (x^7*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/7

Rule 6221

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcT
anh[(c*x)/Sqrt[a + b*x^2]])/(d*(m + 1)), x] - Dist[c/(d*(m + 1)), Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /;
FreeQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^6 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{7} \sqrt{e} \int \frac{x^7}{\sqrt{d+e x^2}} \, dx\\ &=\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{14} \sqrt{e} \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{d+e x}} \, dx,x,x^2\right )\\ &=\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{14} \sqrt{e} \operatorname{Subst}\left (\int \left (-\frac{d^3}{e^3 \sqrt{d+e x}}+\frac{3 d^2 \sqrt{d+e x}}{e^3}-\frac{3 d (d+e x)^{3/2}}{e^3}+\frac{(d+e x)^{5/2}}{e^3}\right ) \, dx,x,x^2\right )\\ &=\frac{d^3 \sqrt{d+e x^2}}{7 e^{7/2}}-\frac{d^2 \left (d+e x^2\right )^{3/2}}{7 e^{7/2}}+\frac{3 d \left (d+e x^2\right )^{5/2}}{35 e^{7/2}}-\frac{\left (d+e x^2\right )^{7/2}}{49 e^{7/2}}+\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0553371, size = 79, normalized size = 0.69 \[ \frac{\sqrt{d+e x^2} \left (-8 d^2 e x^2+16 d^3+6 d e^2 x^4-5 e^3 x^6\right )}{245 e^{7/2}}+\frac{1}{7} x^7 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(Sqrt[d + e*x^2]*(16*d^3 - 8*d^2*e*x^2 + 6*d*e^2*x^4 - 5*e^3*x^6))/(245*e^(7/2)) + (x^7*ArcTanh[(Sqrt[e]*x)/Sq
rt[d + e*x^2]])/7

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Maple [B]  time = 0.037, size = 224, normalized size = 2. \begin{align*}{\frac{{x}^{7}}{7}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{1}{7\,d}{e}^{{\frac{3}{2}}} \left ({\frac{{x}^{8}}{9\,e}\sqrt{e{x}^{2}+d}}-{\frac{8\,d}{9\,e} \left ({\frac{{x}^{6}}{7\,e}\sqrt{e{x}^{2}+d}}-{\frac{6\,d}{7\,e} \left ({\frac{{x}^{4}}{5\,e}\sqrt{e{x}^{2}+d}}-{\frac{4\,d}{5\,e} \left ({\frac{{x}^{2}}{3\,e}\sqrt{e{x}^{2}+d}}-{\frac{2\,d}{3\,{e}^{2}}\sqrt{e{x}^{2}+d}} \right ) } \right ) } \right ) } \right ) }-{\frac{1}{7\,d}\sqrt{e} \left ({\frac{{x}^{6}}{9\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{2\,d}{3\,e} \left ({\frac{{x}^{4}}{7\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{4\,d}{7\,e} \left ({\frac{{x}^{2}}{5\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{2\,d}{15\,{e}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

1/7*x^7*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/7*e^(3/2)/d*(1/9*x^8/e*(e*x^2+d)^(1/2)-8/9*d/e*(1/7*x^6/e*(e*x^2+
d)^(1/2)-6/7*d/e*(1/5*x^4/e*(e*x^2+d)^(1/2)-4/5*d/e*(1/3*x^2/e*(e*x^2+d)^(1/2)-2/3*d/e^2*(e*x^2+d)^(1/2)))))-1
/7*e^(1/2)/d*(1/9*x^6*(e*x^2+d)^(3/2)/e-2/3*d/e*(1/7*x^4*(e*x^2+d)^(3/2)/e-4/7*d/e*(1/5*x^2*(e*x^2+d)^(3/2)/e-
2/15*d/e^2*(e*x^2+d)^(3/2))))

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Maxima [A]  time = 0.986734, size = 209, normalized size = 1.83 \begin{align*} \frac{1}{7} \, x^{7} \operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right ) - \frac{35 \,{\left (e x^{2} + d\right )}^{\frac{9}{2}} - 135 \,{\left (e x^{2} + d\right )}^{\frac{7}{2}} d + 189 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{3}}{2205 \, d e^{\frac{7}{2}}} + \frac{35 \,{\left (e x^{2} + d\right )}^{\frac{9}{2}} - 180 \,{\left (e x^{2} + d\right )}^{\frac{7}{2}} d + 378 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} d^{2} - 420 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{3} + 315 \, \sqrt{e x^{2} + d} d^{4}}{2205 \, d e^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/7*x^7*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)) - 1/2205*(35*(e*x^2 + d)^(9/2) - 135*(e*x^2 + d)^(7/2)*d + 189*(e*x
^2 + d)^(5/2)*d^2 - 105*(e*x^2 + d)^(3/2)*d^3)/(d*e^(7/2)) + 1/2205*(35*(e*x^2 + d)^(9/2) - 180*(e*x^2 + d)^(7
/2)*d + 378*(e*x^2 + d)^(5/2)*d^2 - 420*(e*x^2 + d)^(3/2)*d^3 + 315*sqrt(e*x^2 + d)*d^4)/(d*e^(7/2))

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Fricas [A]  time = 2.20161, size = 205, normalized size = 1.8 \begin{align*} \frac{35 \, e^{4} x^{7} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right ) - 2 \,{\left (5 \, e^{3} x^{6} - 6 \, d e^{2} x^{4} + 8 \, d^{2} e x^{2} - 16 \, d^{3}\right )} \sqrt{e x^{2} + d} \sqrt{e}}{490 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/490*(35*e^4*x^7*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(5*e^3*x^6 - 6*d*e^2*x^4 + 8*d^2*e*x^
2 - 16*d^3)*sqrt(e*x^2 + d)*sqrt(e))/e^4

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Sympy [A]  time = 19.2527, size = 116, normalized size = 1.02 \begin{align*} \begin{cases} \frac{16 d^{3} \sqrt{d + e x^{2}}}{245 e^{\frac{7}{2}}} - \frac{8 d^{2} x^{2} \sqrt{d + e x^{2}}}{245 e^{\frac{5}{2}}} + \frac{6 d x^{4} \sqrt{d + e x^{2}}}{245 e^{\frac{3}{2}}} + \frac{x^{7} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{7} - \frac{x^{6} \sqrt{d + e x^{2}}}{49 \sqrt{e}} & \text{for}\: e \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((16*d**3*sqrt(d + e*x**2)/(245*e**(7/2)) - 8*d**2*x**2*sqrt(d + e*x**2)/(245*e**(5/2)) + 6*d*x**4*sq
rt(d + e*x**2)/(245*e**(3/2)) + x**7*atanh(sqrt(e)*x/sqrt(d + e*x**2))/7 - x**6*sqrt(d + e*x**2)/(49*sqrt(e)),
Ne(e, 0)), (0, True))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError